Question

At a call center, calls come in at an average rate of four calls per minute. Assume that the time elapsed from one call to the next has the exponential distribution, and that the times between calls are independent.

a. Find the average time between two successive calls.

b. Find the probability that after a call is received, the next call occurs in less than ten seconds.

c. Find the probability that less than five calls occur within a minute.

d. Find the probability that more than 40 calls occur in an eight-minute period.

e. Find a 95% confidence interval for the number of calls in a minute.

Answer 1

(a).

On average there are four calls occur per minute, so 15 seconds, or 15/60= 0.25 minutes occur between successive calls on average.

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(b)

Let T = time elapsed between calls. From part a, μ=0.25, so m = 1/0.25 = 4. Thus, T ~ Exp(4). The cumulative distribution function is P(T < t) = 1 – e^–4t. The probability that the next call occurs in less than ten seconds (ten seconds = 1/6 minute) is P(T < 1/6) = 1 – exp(−4/6) = 0.4866.

T(t) = Exp(4) = exponential distribution = exp(-4t)

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(c).

Keep in mind that X must be a whole number, so P(X < 5) = P(X ≤ 4).

To compute this, we could take P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Using technology, we see that P(X ≤ 4) = 0.6288.

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(d).

Let Y = the number of calls that occur during an eight minute period.

Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period.

Hence, Y ∼ Poisson(32). Therefore, P(Y > 40) = 1 – P (Y ≤ 40) = 1 – 0.9294 = 0.0707.

Poisson(32) = Y(t) = (32^t * exp(-32) ) / t!

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(e).

We have four calls per minute, thus estimated standard deviation is S = = 2

So the 95% confidence interval is =

so 95% confidence interval for the number of calls in a minute is (0.08, 7.92)

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